The given limit is:

$\lim_{x \to 0} \frac{\sin(5x)}{\sin(10x)}$

Solution:

1. Direct Substitution: We cannot directly substitute $x = 0$ because both $\sin(5x)$ and $\sin(10x)$ will be 0 at $x = 0$, leading to the indeterminate form $\frac{0}{0}$.

2. Apply Limit Properties: To resolve this, we can use the standard limit property: $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$ for any constant $k$.

3. Rewrite the Expression: We manipulate the limit expression by multiplying both the numerator and denominator by $x$: $\lim_{x \to 0} \frac{\sin(5x)}{\sin(10x)} = \lim_{x \to 0} \frac{\frac{\sin(5x)}{x}}{\frac{\sin(10x)}{x}}$ Now, we can use the limit property: $\lim_{x \to 0} \frac{\sin(5x)}{5x} \times \frac{10x}{\sin(10x)} = \frac{5}{10} = \frac{1}{2}$

Thus, the value of the limit is: $\boxed{\frac{1}{2}}$

Would you like further clarification or details?

Here are 5 related questions:

1. How would this limit change if the functions involved were $\sin^2(5x)$ and $\sin^2(10x)$?
2. What happens if we change the angle coefficients to different constants, say $\sin(3x)$ and $\sin(7x)$?
3. How do you calculate limits using L'Hopital's Rule in cases of indeterminate forms like $\frac{0}{0}$?
4. What are some other common trigonometric limits that are frequently encountered?
5. Can you explain the intuition behind the standard limit property $\lim_{x \to 0} \frac{\sin(kx)}{kx} = 1$?

Tip:

When dealing with limits involving trigonometric functions, the small-angle approximation $\sin(x) \approx x$ near zero is incredibly useful for resolving indeterminate forms.